Let $K$ be an infinite countable subset of Euclidean space $E$ such any point of $E$ is within distance 1 of some point of $K$. In the language of John Roe's "coarse geometry", this implies that $K$ is coarsely equivalent to $E$. Let us also assume that $K$ is uniformly discrete, i.e. that the distance between distinct points of $K$ is bounded from below by a positive constant. Consider the Vietoris-Rips complex of $K$. It depends on a parameter delta (a positive number). What can one say about the homology of this complex for large delta? Is it true that it stabilizes for sufficiently large delta? Is the limit isomorphic to the homology of $E$ (that is, trivial in degree greater than $0$)? There is also a "Borel-Moore" version of this question, where one allows locally-finite but not necessarily finite chains.

If you strengthen the uniform discreteness assumption a little you can ensure that the Rips complex of any such $K$ is contractible for large enough $\delta$. The strengthened assumption I have in mind is to require the set of distances between points in $K$ to form a discrete subset of $\mathbb{R}$ (so the distances don't accumulate anywhere in $\mathbb{R}$ - your requirement is that they just don't accumulate at $0$). Then contractibility of $Rips_\delta(K)$ for large $\delta$ follows from Example 4.7 in https://arxiv.org/abs/1812.10976v2 using $r=1$.

This condition about the set of distances being discrete is equivalent to your condition about non-zero distances being bounded away from zero in some special cases. For example they're equivalent if $K$ is the orbit of a finite subset under the action of some group of isometries of $E$. But I don't know if the kinds of $K$ you have in mind arise that way. For the record an example that satisfies your condition but not mine is $E=\mathbb{R}$ and $K=\{m\in\mathbb{Z}\mid m\le 1\} \cup \{n+\frac{1}{n}\mid n\in\mathbb{N}\}$.

this means that $K$ is coarsely equivalent to $E$". No: this means that the embedding of $K$ into $E$ is a coarse equivalence. (It's maybe true that $E$ is not coarsely equivalent to any non-cobounded subset, but if so that's certainly not a definition, but a theorem.) $\endgroup$ – YCor Aug 15 '19 at 5:58